The proof is similar to the duality line case. Therefore we only detail the parts of the argument that are different. We use the same decomposition, and keep the notations of the previous subsection.
To bound the singular part, we write (see [1]} for a similar reasoning in the euclidean case) using Plancherel's theorem and Theorem REF  that
\(\Vert 2^k \Psi (2^k(\lambda -\sqrt{\tau })) \widetilde{f}(\lambda ,\omega ) \Vert _{L^2}^2 & =2^{2k} \int _{0}^{\infty } \vert \Psi (2^k(\lambda -\sqrt{\tau })) \vert ^2 \Vert R_{\lambda } f \Vert _{L^2_{\omega }}^2 d\lambda \\& \lesssim 2^k \Vert f \Vert _{L^{p^{\prime }}}^2 \left\lbrace  \begin{array}{ll}\tau ^{-\frac{d+1}{2}+\frac{d}{p^{\prime }}} & \mbox{if $p \ge p_{ST} = \frac{2(d+1)}{d-1}$} \\\tau ^{\frac{d-1}{2}\big (\frac{1}{p^{\prime }}-\frac{1}{2} \big )} & \mbox{if $2 < p \le p_{ST}$}\end{array} \right. .\) 
